I am given this finite field. $$K=\mathbb{Z_3}[x]/p$$ with $$p(x)=x^2+1$$
In my textbook, i see that the elements of this field are:
$$\{0, 1, 2, x, x +1, x+2, 2x, 2x + 1, 2x + 2\}$$
I dont understand how these come. I know that $0, 1, 2$ are elements of $\mathbb{Z_3}$, but where is $x+$ is coming from?
Note that $\Bbb{Z}_3[x]$ is the set of all polynomials whose coefficients are elements of $\Bbb{Z}_3$. So this includes things like $0$, $x + 1$, $x^2 + 2x + 1$, $x^{100}$, and so on. When we take the quotient, we're essentially dividing all these polynomials by $x^2 + 1$ and considering elements to be the same as their remainders. So the elements of the quotient ring $\Bbb{Z}_3[x]$ are precisely the polynomials which are remainders upon division by $x^2 + 1$.
So the polynomial $x$, when divided by $x^2 + 1$ looks like
$$x = 0(x^2 + 1) + x$$
So $x$ is a remainder, and so appears in the list. On the other hand,
$$x^3 + 2x^2 + 2x + 1 = (x + 2)(x^2 + 1) + x + 2$$
So we see that $x + 1$ is a remainder as well, and must appear on our list (more directly, $x^2 + 1$ is its own remainder upon division).
Note that if $f$ is any polynomial, then $f/(x^2 + 1)$ will have a remainder which is either constant or degree $1$, so the quotient ring will actually be quite small (prove this!).